This presentation covers the following points:-
Requirements of gear box for Machine Tool Application
Basic Considerations in the Design of Multi-Speed Gear Box
Determination of Variable Speed Range
1. Arithmetic Progression (AP)
2. Geometric Progression (GP)
3. Harmonic Progression (HP)
Selection of Range Ratio (RN)
Selection of GP Ratio (β )
Structure Formula
Structure Diagram
Ray Diagram
Rules and Guidelines For Gear Box Layout
Some Gear Box Layout
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Design of gear box for Machine Tool Application (3 stage & 12 speed ) by Sagar Dhotare
1. Design of Gear Box
for
M/c Tool Application
(3-Stage & 12 Speeds)
Presented by
Prof. Sagar A. Dhotare,
Assistant Professor,
ViMEET, Khalapur
2. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
Introduction of Gear Box
for Machine Tool Application
β’ Gear box is a mechanism used for transmitting a power from prime mover to the machine with change in torque
and speed.
β’ Machine tool expect to perform various operations such as turning, facing, milling, boring etc. For best
performance of each operation is depend on given cutting speed, feed and depth of cut.
β’ Based on the diameter to be machined of the job/workpiece, different spindle speeds are necessary to determine
optimum cutting speed and feed rates.
β’ Different possible spindle speeds for different diameter of the job need to be generated by multi-speed gear box.
This multi-speed gear box is used for stepped regulation of speeds.
Types of Gear Boxes
1. Sliding Mesh Gear Box.
2. Constant Mesh Gear Box.
3. Synchromesh Gear Box.
3. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
1. The speed range between to increments should be 10 to 15%. A large range, would further reduce the
optimum use of the tool.
2. A no. of speed ratio should be available to the operator without stopping the machine.
3. The operator should able to change speeds without going through any intermediate sequence.
4. Due to compact system, axial length of the gear train is the least, if the gear cluster are made sliding.
But, the sliding gear cluster should not have more than three gears.
5. Only the gears which are used to achieve the gear ratio should be rotating, whereas other gears of the
gear box should not be engaged.
6. The no. of shaft, gears and operating levers should be minimum. Lesser the no. of components,
machine reliability is higher.
7. Control unit should be designed ergonomically. It should be centralised and should have all possible
controls at one place.
4. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
The ideal speed range in a machine tool would be the one were there would be infinity no. of speed ratios.
This would allow an optimum operation of a machine for any job diameter. Such a speed regulation would be called as
a Stepless Regulation.
Higher the speed ratio available, higher would be the cost of the gear box. Hence, from the cost and utility
point of view, the optimum gear ratios should be available to get the different speed ranges between the maximum and
minimum speed. These intermediate discrete values can be obtained by different laws.
Let Nmin. and Nmax. be the minimum and maximum spindle speeds to be achieved in βzβ no. of intermediate
speed steps.
These steps can be arranged by the following laws :
1. Arithmetic Progression (AP)
2. Geometric Progression (GP)
3. Harmonic Progression (HP)
5. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
In Arithmetic Progression (AP) the difference between two adjacent speed steps is constant.
Various intermediate speeds can be found using following sequence :
π΅π = π΅πππ
π΅π = π΅π + π = π΅πππ + π
Where, a - Difference between two adjacent speed steps,
which is constant.
π΅π = π΅π + π = π΅πππ + ππ
π΅π = π΅πβπ + π = π΅πππ + (π β π)π
π΅π+π = π΅π + π = π΅πππ + ππ
π΅π = π΅πβπ + π = π΅πππ + π β π π = π΅πππ
π =
π΅πππ β π΅πππ
π β π
Where, z β no. of steps.
6. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
π«π =
ππππ π½
π π΅π
Considering any two intermediate speed ranges Nk and Nk+1. Corresponding to these, the diameters
that can be machined would be as follows,
Upper Limit in mm
π«π+π =
ππππ π½
π π΅π+π
Lower Limit in mm
The diameter range that can be accommodated at a particular speed would be as follows,
βπ«π = π«π β π«πβπ =
ππππ π½
π
π
π΅π
β
π
π΅π+π
7. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
Example of AP :-
Considering a gear box to be designed for a minimum and maximum speed of 40rpm to 400rpm in
6 steps. Consider the optimum velocity for the tool be 20mmin.
π΅π = π΅πππ = 40 rpm
π΅π = π΅π = π΅πππ = 400 rpm
z = 6
V = 20 mmin.
π =
π΅πππ β π΅πππ
πβπ
=
400 β40
6β1
= 72rpm
Let the gear ratio between adjacent speeds be Ο.
Hence,
Οπ
=
ππ+1
ππ
Sr.
No.
N (rpm) Οπ =
ππ+1
ππ
π·π βπ·π
1 40 (112/40) = 2.8 159.15 102.3
2 40 + 72 = 112 (184/112) = 1.64 56.84 22.2
3 112 + 72 = 184 (256/184) = 1.4 34.6 9.73
4 184 + 72 = 256 (328/256) = 1.28 24.87 5.46
5 256 + 72 = 328 (400/328) = 1.22 19.4 3.5
6 328 + 72 = 400 - 15.9
The diameter and ranges with their gear ratio are
calculated as follows(table),
Note : In the Arithmetic Progression, it is seen that at low speed steps, more speed ranges are desired than
what is available. Where as at high speed steps, some of the speed ranges available are not necessary.
8. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
In Geometric Progression (GP) the difference between two consecutive speed ranges is same.
Considering the similar example as discussed for AP
π΅π = π΅πππ
π΅π = π΅πππ
As per definition of GP
π2
π1
=
π3
π2
=
π4
π3
= β¦ =
ππ+1
ππ
=
ππ§β1
ππ§
= Ο
ππ§
π1
=
ππππ₯
ππππ
= Ο
Ο =
π΅πππ
π΅πππ
π
(πβπ)
9. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
Example of GP :-
Considering a gear box to be designed for a minimum and maximum speed of 40rpm to 400rpm in
6 steps. Consider the optimum velocity for the tool be 20mmin.
π΅π = π΅πππ = 40 rpm
π΅π = π΅π = π΅πππ = 400 rpm
z = 6
V = 20 mmin.
Sr.
No. N (rpm) Οπ =
ππ+1
ππ
π·π βπ·π
1 40 1.585 159.15 58.73
2 40 X 1.585 = 63.4 1.585 100.4 37.05
3 63.4 X 1.585 = 100.5 1.585 63.33 23.4
4 100.5 X 1.585 = 159.6 1.585 40 14.78
5 159.6 X 1.585 = 252.5 1.585 25.22 9.3
6 252.5 X 1.585 = 400 - 15.9
The diameter and ranges with their gear ratio are
calculated as follows(table),
Note : In the Geometric Progression, it is seen that at low speed steps, still more speed ranges are desired
than what is available. Where as at high speed steps, some of the speed ranges available are not necessary.
Ο =
π΅πππ
π΅πππ
π
(πβπ)
Ο =
400
40
1
(6β1)
Ο = 1.585
10. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
In Harmonic Progression (HP) the diameter changes between two consecutive speed ranges remains same.
Considering the similar example as discussed for AP and GP
βπ«π =
ππππ π½
π
π
π΅π
β
π
π΅π+π
= Constant
π
π΅π
β
π
π΅π+π
= C
π
π΅π+π
=
π
π΅π
- C =
π βπͺπ΅π
π΅π
π΅π+π =
π΅π
πβπͺπ΅π
π΅π = π΅πππ
π΅π =
π΅π
π β πͺπ΅π
π΅π =
π΅π
π β ππͺπ΅π
π΅π =
π΅π
π β π β π πͺπ΅π
π΅π =
π΅πππ
π β π β π πͺπ΅πππ
π΅π =
π΅πβπ
π β π β π πͺπ΅π
π΅π =
π΅πππ
π β π β π πͺπ΅πππ
= π΅πππ
πͺ =
π΅πππ β π΅πππ
π β π π΅πππ π΅πππ
Οπ
=
π΅π+π
π΅π
=
π
πβπͺπ΅π
11. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
Example of HP :-
Considering a gear box to be designed for a minimum and maximum speed of 40rpm to 400rpm in
6 steps. Consider the optimum velocity for the tool be 20mmin.
π΅π = π΅πππ = 40 rpm
π΅π = π΅π = π΅πππ = 400 rpm
z = 6
V = 20 mmin.
Sr.
No.
Οπ
=
ππ+1
ππ
π·π βπ·π
1 40 (48.78 / 40) = 1.22 159.15 28.65
2 40
1 β 4.5 Γ 10β3 Γ 40
= 48.78 (62.5 / 48.78) = 1.28 130.5 28.65
3 48.78
1 β 4.5 Γ 10β3 Γ 48.78
= 62.5 (86.96 / 62.5) = 1.39 101.86 28.65
4 62.5
1 β 4.5 Γ 10β3 Γ 62.5
= 86.96 (142.86 / 86.96) = 1.643 73.21 28.65
5 86.96
1 β 4.5 Γ 10β3 Γ 86.96
= 142.86 (400 / 142.86) = 2.8 44.56 28.65
6 142.86
1 β 4.5 Γ 10β3 Γ 142.86
= 400 - 15.9 -
The diameter and ranges with their gear ratio are calculated as follows(table),
Note : In the comparison with AP & GP series, the HP gives a better distribution of speeds in the lower speed
range. At higher speeds, the values are spaced too wide apart which results in non-optimum use of tool.
πͺ =
π΅πππ β π΅πππ
π β π π΅πππ π΅πππ
=
400 β 40
6 β 1 π 400 π 40
πͺ = 4.5 X 10β3
ππ =
ππππ
1 β π β 1 πΆππππ
12. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
Sr.
No. Arithmetic Progression (AP) Geometric Progression (GP) Harmonic Progression (HP)
1
Constant Change in speed at every stage
is constant
Constant gear ratio at every stage
Constant diameter change at every
stage
2
π =
π΅πππ β π΅πππ
π β π
Ο =
π΅πππ
π΅πππ
π
(πβπ) πͺ =
π΅πππ β π΅πππ
π β π π΅πππ π΅πππ
3
π΅π = π΅πππ + (π β π)π π΅π = π΅πππ Ο(πβπ)
π΅π =
π΅πππ
π β π β π πͺπ΅πππ
4 Low speed zone requires much more steps
for optimised machining.
Low speed zone, better than AP, but still
needs few more steps for optimised
machining.
High speed zone requires much more
steps for optimised machining.
5 Not based on a preferred no. of series. Based on preferred no. of series Not based on a preferred no. of series.
A. Constant loss of Economic cutting speed in the whole rpm range
B. Better gear box design feature
Advantages of GP
13. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
π π =
ππππ₯
ππππ
ππππ₯ =
1000 Γ π
πππ₯
π π·πππ
ππππ =
1000 Γ ππππ
π π·πππ₯
π π =
π
πππ₯
ππππ
Γ
π·πππ₯
π·πππ
= π π£ Γ π π·
The ratio π π is defind as ratio of maximum spindle speed to minimum spindle speed
β’ Fixing the limit on higher cutting speeds results in loss of productivity.
β’ Machining operation can be done at higher speeds, but they are not available with machine.
β’ At the same time very high limits of upper cutting speed is not practical, nor economical from machine
building point of view.
β’ The lower cutting speed is determined by the considering that below a particular cutting speed, the tool life
reduces.
β’ In general purpose machine such as lathes , boring and milling machines, the magnitude of π π is large to
account of different of jobs that can be machined on them.
14. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
To find the geometric progression ratio β we refer Preferred Number Series. This is because following resons;
1. Each size is larger than the preceding size by a fixed percentage.
2. Small size will differ from each other by small amounts and large sizes will differ from each other by larger
amount
The series of numbers have been standardised in order that the various
sizes of a series can be determined in an orderly fashion and these are
called as Preferred Number which shows in tabular form
Generally β ππ ππ πππππ ππππ 1 π‘π 2
Ο =
π΅πππ
π΅πππ
π
(πβπ)
= πΉπ΅
π
(πβπ)
15. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
The values obtained from above equation may not be a whole no. and hence it should be rounded off to the
nearest whole no. with a preferred given to the no.
π§ =
log(π π β )
log β
Range of z z Rounded Value
5 to 6.49 6 (2 Γ 3)
7 to 8.49 8 (2 Γ 2 Γ 2) = 23
8.5 to 10 9 (3 Γ 3) = 32
10 to 13 12 (2 Γ 2 Γ 3) = 22
Γ 3
13 to 16 16 (2 Γ 2 Γ 2 Γ 2) = 24
Generally the value of z are selected on the basis of following equation,
π§ = 2π1 Γ 3π2
π1- No. stages having two speed steps
π2- No. stages having three speed steps
Total no. of steps (n) = π1+ π2
Standard Values of z and n
z πππ Γ πππ ππ ππ n
2 21 Γ 30 1 0 1
4 22 Γ 30 2 0 2
6 21
Γ 31
1 1 2
8 23
Γ 30
3 0 3
9 20 Γ 33 0 2 2
12 21
Γ 32
2 1 3
16. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
It can be defind as an expression which gives the distribution of the no. of transmission stages and the
difference between the adjacent speed step in each of these stages.
Following is the procedure to calculate Structural Formula,
Consider βzβ be the total no. speed steps
Let π1, π2,ππ are the no. of speed steps in first, second and nth stage of transmission.
z = π1 Γ π2 Γ π3 Γ β¦β¦β¦. Γ ππ where n β no. of stages of the gear box
Consider a case where we analyse the difference between two adjacent speeds βxβ in a given stage, x is referred
to as the characteristics of transmission stage.
Consider any two adjacent speeds available in this stage. If the difference between these two speeds is 1, then
x = 1, where as if the value is 2, then x = 2. Thus for each stage there will be particular value of βxβ.
As the speed of the o/p shaft is GP, there must be one transmission stage which would give x = 1. This group is called as
Main Transmission group and has a progression ratio of β π = β π = β .
The next group would have ππ = ππ and the progression ratio of β ππ = β ππ
Next transmission group, ππ = ππππ and the progression ratio of β ππ = β ππππ
17. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
The values of π₯1, π₯2, π₯3, β¦ . . , π₯π can be calculated as,
π₯1 = 1
π₯2 = π1
π₯3 = π1 π2
.
.
π₯π = π1 Γ π2 Γ π3 Γ β― Γ ππβ1
So final structural formula give as,
π = ππ(ππ) ππ(ππ) ππ(ππ) β¦β¦β¦β¦.. ππ(ππ)
18. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
The representation of structural formula in the form of special graph is called as Structure Diagram.
Which describes the kinematic structure of the gear box.
z = 2(1) 2(2) z = 2(2) 2(1)
Rules for drawing structure diagram:-
1. If the no. of stages in a gear box are βnβ, draw
βn+1β vertical lines spaced at a convenient
distance.
2. 1st vertical line indicate I/P shaft and last line
indicates O/P shaft. Other vertical lines
indicates intermediate shafts which forms
transmission groups or stages of the gear
box.
3. βmβ horizontal lines are drawn to represent
total no. of speeds steps of the gear box.
Spacing between the horizontal lines is taken
as logβ .
4. Using structural formula
π = ππ(ππ) ππ(ππ).. ππ(ππ)
Plot π1(π₯1) π2(π₯2) etc
π1 = 2 π₯1 = 1 π₯2 = π1 = 2
Z=4 n=2
19. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
Flow of transmission to get the 4 speeds
Table shows Preferred structural formula
Few Structure diagram combinations
20. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
To Designer what information get from Structural Diagram :-
1. The no. of shafts in the gear box.
2. The no. of gears on each shaft. The approximate no. of gears would be found by following equation:
πΊ = 2 Γ π1 + π2+. . . . +ππ
3. The order of changing transmissions at each stage to get the required spindle speed.
4. The transmission range and the characteristics of each group
Characteristics of a structure diagram are as follows:
1. Each line must connect another follow-up line till the final O/P shaft is reached.
2. At any stage the arrow must terminate at one and only point. However at any input there can be more than one
arrow. Generally this no. is limited to a maximum of three.
3. On the final shaft, all the points must be connected by arrows.
4. No arrow should go beyond the bounds of the speed required speeds at the final O/P shaft.
21. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
Structural diagram gives the information about the transmission flow only not about the speed. To
determine the speeds of various gears, the ray diagram and speed diagram are drawn. This diagram allows
the designer to read the speeds of all shafts gears and finally spindle speeds.
Following are the steps to draw the ray and speed diagram:-
1. If the no. of steps in a gear box is βzβ, and the no. of stages of the gear box is βnβ, then draw βn+2β vertical lines at
a convenient distance. The one extra line as compared to the structural diagram would represent the motor shaft.
2. If the I/P speed from motor is less then the maximum speed, draw βzβ horizontal lines, where as if the I/P speed
from motor is more then the maximum speed, draw as many horizontal lines as necessary to locate the motor
speed on the diagram . The spacing between the horizontal lines take as log β , like structure diagram. Thus,
If the ππ < ππππ₯ , z no. of Horizontal lines and
If the ππ > ππππ₯ , z no. of Horizontal lines + additional lines to locate ππ on the chart.
3. Using the value GP ratio 'β β²
, mark the spindle speeds on the vertical shaft. Mark the various spindle spedds
starting with minimum speed at the lowermost line.
4. Draw the ray depicting transmission between the last shft and the shaft preceding it. The ray are drawn for the
lowest speed of the last shaft.
22. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
23. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
After completion of structure and speed diagram, a rough layout of the gears should be drawn. This layout is
used to determine the no. of teeth on each gear based on speed calculation referring speed diagram. The gears,
shafts and other component are designed on the basis of strength consideration.
While preparing the gear layout, the following rules should be consider :
1. Minimum Number of Teeth on Gears.
2. Sum of Number of Teeth on Gear Pairs for Parallel Shafts Should be The same.
3. Check on Percent Deviation of Actual and Theoretical Speeds.
4. The spacing between to adjacent Gears should be proper.
24. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
25. Prof. Sagar A. Dhotare, Assistant Professor, ViMEET, Khalapur.
Thank You